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Financial Maths, STD1 F3 2023 HSC 30

A plumber leases equipment which is valued at $60 000.

The salvage value of the equipment at any time can be calculated using either of the two methods of depreciation shown in the table.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Method of depreciation} \rule[-1ex]{0pt}{0pt} & \textit{Rate of depreciation} \\
\hline
\rule{0pt}{2.5ex} \text{Straight-line method} \rule[-1ex]{0pt}{0pt} & \text{\$3500 per annum} \\
\hline
\rule{0pt}{2.5ex} \text{Declining-balance method} \rule[-1ex]{0pt}{0pt} & \text{12% per annum} \\
\hline
\end{array}

Under which method of depreciation would the salvage value of the equipment be lower at the end of 3 years? Justify your answer with appropriate mathematical calculations.  (3 marks)

Show Answers Only

\(\text{Straight-line method:}\)

\(S\) \(=V_0-Dn\)  
  \(=60\ 000-3500\times 3\)  
  \(=$49\ 500\)  

 
\(\text{Declining-balance method:}\)

\(S\) \(=V_0(1-r)^n\)  
  \(=60\ 000(1-0.12)^3\)  
  \(=60\ 000(0.88)^3\)  
  \(=$40\ 888.32\)  

 
\(\text{Salvage value is lower for the declining-balance method.}\)

Show Worked Solution

\(\text{Straight-line method:}\)

\(S\) \(=V_0-Dn\)  
  \(=60\ 000-3500\times 3\)  
  \(=$49\ 500\)  

 
\(\text{Declining-balance method:}\)

\(S\) \(=V_0(1-r)^n\)  
  \(=60\ 000(1-0.12)^3\)  
  \(=60\ 000(0.88)^3\)  
  \(=$40\ 888.32\)  

 
\(\text{Salvage value is lower for the}\)

\(\text{declining-balance method.}\)


♦ Mean mark 48%.

Measurement, STD1 M3 2023 HSC 29

The diagram shows the location of three places \(X\), \(Y\) and \(C\).

\(Y\) is on a bearing of 120° and 15 km from \(X\).

\(C\) is 40 km from \(X\) and lies due west of \(Y\).

\(P\) lies on the line joining \(C\) and \(Y\) and is due south of \(X\).
  

  1. Find the distance from \(X\) to \(P\).  (2 marks)

  2. What is the bearing of \(C\) from \(X\), to the nearest degree?  (2 marks)

Show Answers Only
  1. \(7.5\ \text{km}\)
  2. \(259^{\circ}\)
Show Worked Solution

a.    \(\text{In}\ \Delta XPY:\)

\(\angle PXY=180-120=60^{\circ}\)

\(\cos 60^{\circ}\) \(=\dfrac{XP}{15}\)  
\(XP\) \(=15\times \cos 60^{\circ}\)  
  \(=7.5\ \text{km}\)  

♦♦ Mean mark (a) 24%.

b.    \(\text{In}\ \Delta XPC:\)

\(\text{Let}\ \theta = \angle CXP\)

\(\cos \theta\) \(=\dfrac{7.5}{40}\)  
\(\theta\) \(=\cos^{-1} \Big(\dfrac{7.5}{40}\Big)\)  
  \(=79.193…\)  
  \(=79^{\circ}\ \text{(nearest degree)}\)  

 

\(\text{Bearing}\ C\ \text{from}\ X\) \(=180+79\)  
  \(=259^{\circ}\)  

♦♦♦♦ Mean mark (b) 9%.

Algebra, STD1 A3 2023 HSC 26

Electricity provider \(A\) charges 25 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $40.

  1. Complete the table showing Provider \(A\) 's monthly charges for different levels of electricity usage.  (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ 0\ \ \ \ \  & \ \ \ 400\ \ \  & \ \ 1000\ \  \\
\hline
\rule{0pt}{2.5ex} \text{Monthly charge (\$)} \rule[-1ex]{0pt}{0pt} & 40 & & 290 \\
\hline
\end{array}

Provider \(B\) charges 35 cents per kWh, with no fixed monthly charge. The graph shows how Provider \(B\) 's charges vary with the amount of electricity used in a month.

  1. On the grid on above, graph Provider A's charges from the table in part (a).  (1 mark)
  2. Use the two graphs to determine the number of kilowatt hours per month for which Provider \(A\) and Provider \(B\) charge the same amount.  (1 mark)

  3. A customer uses an average of 800 kWh per month.
  4. Which provider, \(A\) or \(B\), would be the cheaper option and by how much?  (2 marks)

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a.   

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ 0\ \ \ \ \  & \ \ \ 400\ \ \  & \ \ 1000\ \  \\
\hline
\rule{0pt}{2.5ex} \text{Monthly charge (\$)} \rule[-1ex]{0pt}{0pt} & 40 & \textbf{140} & 290 \\
\hline
\end{array}

b.

c.    \(400\text{ kWh}\)

d.    \(A\text{ is cheaper by }$40.\)

Show Worked Solution

a.   

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ 0\ \ \ \ \  & \ \ \ 400\ \ \  & \ \ 1000\ \  \\
\hline
\rule{0pt}{2.5ex} \text{Monthly charge (\$)} \rule[-1ex]{0pt}{0pt} & 40 & \textbf{140} & 290 \\
\hline
\end{array}

b.   


♦♦ Mean mark (b) 40%.

c.    \(\text{Same charge when Provider }A = \text{Provider } B \text{ i.e. where the lines intersect}\)

\(=400\text{ kWh (see graph above)}\)

d.    \(\text{When kWh}= 800, \ \ A=$240 \text{ and } B=$280\)

\(\therefore A\text{ is cheaper by }$40.\)

Networks, STD1 N2 2023 HSC 15

The table shows some of the flight distances (rounded to the nearest 10 km between various Australian cities.

  1. Use the information in the table to complete the network diagram where the edges are labelled with distances.  (2 marks)
     


 

  1. Mahsa wants to travel from Hobart to Darwin. She wants to change planes only once.
  2. Using the network diagram, calculate how many kilometres she will travel by plane.  (1 mark)

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a.  

b.    \(4190\text{ km}\)

Show Worked Solution

a.

b.    \(\text{Changing planes only once}\Rightarrow H\rightarrow S\rightarrow D\)

\(\text{Kilometres travelled}=1040+3150=4190\text{ km}\)

Measurement, STD1 M5 2023 HSC 9 MC

A bag contains 150 jelly beans. Some of them are red and the rest are blue. The ratio of red to blue jelly beans is 2 : 3.

Sophie eats 10 of each colour.

What is the new ratio of red to blue jelly beans?

  1. \(2 : 3\)
  2. \(4 : 9\)
  3. \(5 : 8\)
  4. \(11 : 17\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Original ratio}\ = 2:3\)

\(\text{5 parts = 150}\ \Rightarrow\ \text{1 part}\ = 150/5=30\)

\(\text{Original ratio (by number)}\ = 2 \times 30:3 \times 30 = 60:90\)

\(\text{After eating 10 of each colour:}\)

\(\text{Ratio}\ = 50:80 = 5:8\)

\(\Rightarrow C\)

♦♦ Mean mark 34%.

Financial Maths, STD1 F1 2023 HSC 7 MC

An item was purchased for a price of \($880\), including \(10\%\) GST.

What is the amount of GST included in the price?

  1. \($8.00\)
  2. \($8.80\)
  3. \($80.00\)
  4. \($88.00\)
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\(C\)

Show Worked Solution

\(\text{Let}\ C =\text{Original cost}\)

\(C+0.1 \times C\) \(=880\)  
\(1.1C\) \(=880\)  
\(C\) \(=\dfrac{880}{1.1}\)  
  \(=$800\)  

 
\(\therefore \text{GST}=800\times 0.1=$80\)

\(\Rightarrow C\)

♦ Mean mark 15%.

Financial Maths, STD1 F1 2023 HSC 6 MC

A delivery truck was valued at \($65 000\) when new. The value of the truck depreciates at a rate of \(22\) cents per kilometre travelled.

What is the value of the truck after it has travelled a total distance of \(132 600\) km?

  1. \($35 828\)
  2. \($29 172\)
  3. \($14 872\)
  4. \($14 300\)
Show Answers Only

\(A\)

Show Worked Solution
\(\text{Value}\) \(=65\ 000-132\ 600\times\frac{22}{100}\)
  \(=$35\ 828\)

\(\Rightarrow A\)